Which integral gives the length of the graph of $y=(3x-1)^2$ between $x=0$ and $x=1$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $ \int_0^1\sqrt{1+(6x-2)^2}~dx$ (Choice B) B $ \int_0^1\sqrt{1+(18x-6)^2}~dx$ (Choice C) C $ \int_0^1\sqrt{18x-5}~dx$ (Choice D) D $ \int_0^1\sqrt{6x-1}~dx$
Solution: Recall that the formula for arc length of $~f(x)~$ over $~[a, b]~$ is $ L=\int_a^b\sqrt{1+\big[f\,^\prime(x)\big]^2}~dx\,$. First calculate $~f\,^\prime(x)\,$. $ f\,^\prime(x)=2(3x-1)\cdot3=18x-6$ Next, use the formula above to write the integral expression that gives the arc length in question. $ L=\int_0^1\sqrt{1+(18x-6)^2}~dx$